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Proof step 3 from proof sum-subarray (previous step)

m0 + m +1 ≤ n, n ≤ # ar, ∀ n, ar, m0. n ≤ # ar ∧ m0 + m ≤ n → sum(subarray(ar, 0, n), m0, m) = sum(ar, m0, m) ⊦ sum(subarray(ar, 0, n), m0, m +1) = sum(ar, m0, m +1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
m0 + m < n, m0 + m < # ar, n ≤ # ar, ∀ n, ar, m0. n ≤ # ar ∧ m0 + m ≤ n → sum(subarray(ar, 0, n), m0, m) = sum(ar, m0, m) ⊦ sum(subarray(ar, 0, n), m0, m) = sum(ar, m0, m)

Rule argument

Additional information

Used simplifier rules: sum-succ: n + m < # ar ⊦ sum(ar, n, m + 1) = sum(ar, n, m) + ar[n + m];%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + j = i + i0 ↔ j = i0; (s)% ⊦ n + 0 = n; (s)% ⊦ n0 < n → subarray(ar0, m, n)[n0] = ar0[m + n0]; (s)% ⊦ m < n → n ≠ 0; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m ≤ n → (n < m ↔ false); (s)% ⊦ m + 1 ≤ n ↔ m < n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ # subarray(ar0, m, n) = n; (s)% ⊦ m +1 = m + 1; (s)